On Thu, 29 Nov 2012 09:31:06 -0800 (PST), put
finger to keyboard and composed:

"Wouldn't increasing the capacitance from 0.68uF to 1uF result in a 50%
increase in cooking energy? "

No, that is a common misconception in the business. Do you remember the formula for capacitive reactance ? Apply it and just guess the frequency is over an octave above the sonic range.

The fact is that those caps are not being used as reactive components like in a tuned system, they are being used as coupling caps.

XC = 1/wC = 1 /(2 x pi x 20000 x 1.36 E-6) ~ 6 ohms

My research would suggest that a typical inductance value for the coil
would be of the order of 50uH.

XL = wL = 2 x pi x 20000 x 50 E-6 ~ 6 ohms

So XL = XC, ie resonance.

In that circuit they are effectively in parallel. Ground and the power supply rail are effectively at the same AC potential, so it's not 0.68, it's aready 1.36 uF. That is almost a piece of wire at 20 Khz. Almost, but we are dealing with a quite higher frequency here.

If those 0.68uF capacitors were coupling capacitors, then it wouldn't
matter how large they were. In fact the larger, the better. So let's
assume for the sake of analysis that they were infinitely large. This
means that the voltage at their junction would be constant (Vsupply /
2), irrespective of the induction coil current.

So when the upper IGBT is on, the voltage across the coil would be
Vsupply - Vsupply/2 = Vsupply/2. Similarly, when the bottom IGBT is
on, the coil voltage would be -Vsupply/2. This results in a linearly
increasing coil current, first in one direction and then in the
reverse direction, ie a symmetrical triangular current.

Vsupply/2 = L . dI/dt

I would think that a triangular current would be undesirable.

- Franc Zabkar
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